A man standing on a road has to hold his umbrella at 30 0 with the vertical to keep the rain away. The throws the umbrella and starts running at 10 km/h. He finds that raindrops are hitting his head vertically. Find the speed of raindrops with respect to a. the road, b. the moving man.

Key Concepts Needed

1. Relative Velocity Vector:
   If two objects A and B move with velocities $\vec v_A$ and $\vec v_B$ with respect to the ground, then velocity of A with respect to B is $$\vec v_{A/B}=\vec v_A-\vec v_B$$
2. Rain-man umbrella condition:
   The stick (shaft) of the umbrella is kept parallel to the velocity of rain relative to the man. If the shaft makes 30° with the vertical, then that relative velocity also makes the same 30°.
3. Component Resolution:
   Splitting a vector into horizontal $(x)$ and vertical $(y)$ parts allows easy addition/subtraction.

Logical Chain to Crack the Problem

1. Standing man (initial situation)
   Man’s velocity $\vec v_M = 0$. Hence $$\vec v_{R/M} = \vec v_R - 0 = \vec v_R$$ The direction information: 30° from vertical ⇒ $$\tan 30^{\circ}=\frac{v_{Rx}}{v_{Ry}}$$
2. Running man (second situation)
   Man’s velocity now is $\vec v_M = 10\,\text{km/h}$ (horizontal). He finds drops hit his head vertically, i.e. the horizontal component of $\vec v_{R/M}$ is zero. $$(v_{Rx}-10)=0 \;\Rightarrow\; v_{Rx}=10\,\text{km/h}$$
3. Use both pieces together
   Combine step-1 ratio with step-2 value to get $v_{Ry}$, then magnitude of $\vec v_R$, and finally relative speed when he is running.

That’s all—three straight steps using right-triangle trigonometry and relative velocity formula.

What is happening?

Imagine rain as straight arrows coming down at an angle.

• When the man is standing still, he has to tilt his umbrella so those arrows miss him. The tilt (30° from the straight-down line) tells us how slanted the rain arrows really are.
• When he starts running at 10 km/h, he now carries himself forward. If the rain arrows still point as before, they will now seem to hit him from another direction. He notices that they strike exactly on top of his head (straight down). That means his own forward motion is cancelling the rain’s sideways push.

So we do two detective jobs:

1. Use the 30° tilt to find how slanted (sideways vs. downward) the rain arrows are.
2. Use his 10 km/h run to find the exact size of that sideways arrow.

After that we can rebuild the full rain-arrow speed.

Result:

• Rain speed w.r.t. road ≈ 20 km/h at 30° from vertical.
• Rain speed w.r.t. the running man ≈ 17.3 km/h straight downward.

Step 1 – Set up components for actual rain velocity $\vec v_R$

Let

$$\vec v_R=(v_{Rx})\hat{i}-(v_{Ry})\hat{j}$$

(Downward is taken negative $y$.)

From umbrella tilt when man is standing:

$$\tan 30^{\circ}=\frac{v_{Rx}}{v_{Ry}} \;\Rightarrow\; v_{Rx}=v_{Ry}\tan 30^{\circ}=v_{Ry}\left(\frac{1}{\sqrt{3}}\right)$$

Step 2 – Use running information

Man now runs with $\vec v_M=10\,\text{km/h}\;\hat{i}$. Relative velocity w.r.t. man is

$$\vec v_{R/M}=\vec v_R-\vec v_M=\bigl(v_{Rx}-10\bigr)\hat{i}-v_{Ry}\hat{j}$$

It is felt vertical, so horizontal part = 0:

$$v_{Rx}-10=0 \;\Rightarrow\; v_{Rx}=10\,\text{km/h}$$

Step 3 – Find vertical component $v_{Ry}$

Using the link from Step 1:

$$10=v_{Ry}\tan 30^{\circ}=v_{Ry}\left(\frac{1}{\sqrt{3}}\right) \quad\Longrightarrow\quad v_{Ry}=10\sqrt{3}\approx17.32\,\text{km/h}$$

Step 4 – Magnitude w.r.t. road (ground)

$$|\vec v_R| = \sqrt{v_{Rx}^2+v_{Ry}^2} =\sqrt{10^2+(17.32)^2} =\sqrt{100+300} =\sqrt{400}=20\,\text{km/h}$$

Direction: $30^{\circ}$ to vertical, as given.

Step 5 – Speed w.r.t. running man

With horizontal component zero, only vertical remains:

$$|\vec v_{R/M}| = v_{Ry}=17.32\,\text{km/h}$$

(downward).

Final Answers

• (a) Speed of rain w.r.t. road: $20\,\text{km/h}$, coming $30^{\circ}$ from vertical.
• (b) Speed of rain w.r.t. running man: $17.32\,\text{km/h}$ straight downward.